A 4.0−L sample of a 6.3M NaCl solution is diluted to 50L .
What is the molarity of the diluted solution?
How would I set this up?
mL1 x M1 = mL2 x M2
4.0L x 6.3M = 50L x M2
or you can do it with proportions by
6.3M x 4/50 = ?
A 44.8−mL sample of a 7.0M KNO3 solution is diluted to 1.10L What volume of the diluted solution contains 16.0g of KNO3?
Would I do the same thing for this one or would does it have an extra step it it.
See your other post for the second one.
To set up this problem, you can use the concept of dilution, which is when you add solvent (usually water) to a solution to decrease its concentration. The key idea in dilution is that the moles of solute before and after dilution remain the same.
Here's how you can set up the problem:
1. Identify the given information:
- Initial volume (V1) = 4.0 L
- Initial molarity (M1) = 6.3 M
- Final volume (V2) = 50 L
2. Use the equation for dilution:
M1V1 = M2V2
- M1 is the initial molarity
- V1 is the initial volume (before dilution)
- M2 is the final molarity
- V2 is the final volume (after dilution)
3. Plug in the values into the equation:
(6.3 M)(4.0 L) = M2(50 L)
4. Solve for M2:
M2 = (6.3 M)(4.0 L) / 50 L
M2 = 0.504 M
Therefore, the molarity of the diluted solution is 0.504 M.